3.40.72 \(\int \frac {3 x+\frac {x^3 (-27+18 x-3 x^2)}{e^{16}}}{9 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ -e^{5+3 (-7+\log (x))}+\frac {3}{3-x}+\log (3) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1594, 27, 1586, 1850} \begin {gather*} \frac {3}{3-x}-\frac {x^3}{e^{16}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x + (x^3*(-27 + 18*x - 3*x^2))/E^16)/(9*x - 6*x^2 + x^3),x]

[Out]

3/(3 - x) - x^3/E^16

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x+\frac {x^3 \left (-27+18 x-3 x^2\right )}{e^{16}}}{x \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {3 x+\frac {x^3 \left (-27+18 x-3 x^2\right )}{e^{16}}}{(-3+x)^2 x} \, dx\\ &=\int \frac {3-\frac {27 x^2}{e^{16}}+\frac {18 x^3}{e^{16}}-\frac {3 x^4}{e^{16}}}{(-3+x)^2} \, dx\\ &=\int \left (\frac {3}{(-3+x)^2}-\frac {3 x^2}{e^{16}}\right ) \, dx\\ &=\frac {3}{3-x}-\frac {x^3}{e^{16}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \begin {gather*} \frac {3 \left (9-\frac {e^{16}}{-3+x}-\frac {x^3}{3}\right )}{e^{16}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x + (x^3*(-27 + 18*x - 3*x^2))/E^16)/(9*x - 6*x^2 + x^3),x]

[Out]

(3*(9 - E^16/(-3 + x) - x^3/3))/E^16

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 22, normalized size = 0.92 \begin {gather*} -\frac {{\left (x^{4} - 3 \, x^{3} + 3 \, e^{16}\right )} e^{\left (-16\right )}}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+18*x-27)*exp(3*log(x)-16)+3*x)/(x^3-6*x^2+9*x),x, algorithm="fricas")

[Out]

-(x^4 - 3*x^3 + 3*e^16)*e^(-16)/(x - 3)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 15, normalized size = 0.62 \begin {gather*} -x^{3} e^{\left (-16\right )} - \frac {3}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+18*x-27)*exp(3*log(x)-16)+3*x)/(x^3-6*x^2+9*x),x, algorithm="giac")

[Out]

-x^3*e^(-16) - 3/(x - 3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 16, normalized size = 0.67

method	result	size
risch	\(-{\mathrm e}^{-16} x^{3}-\frac {3}{x -3}\)	\(16\)
default	\(-\frac {3}{x -3}-{\mathrm e}^{3 \ln \relax (x )-16}\)	\(18\)
norman	\(\frac {3 \,{\mathrm e}^{-16} x^{3}-{\mathrm e}^{-16} x^{4}-3}{x -3}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2+18*x-27)*exp(3*ln(x)-16)+3*x)/(x^3-6*x^2+9*x),x,method=_RETURNVERBOSE)

[Out]

-exp(-16)*x^3-3/(x-3)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 15, normalized size = 0.62 \begin {gather*} -x^{3} e^{\left (-16\right )} - \frac {3}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+18*x-27)*exp(3*log(x)-16)+3*x)/(x^3-6*x^2+9*x),x, algorithm="maxima")

[Out]

-x^3*e^(-16) - 3/(x - 3)

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 15, normalized size = 0.62 \begin {gather*} -\frac {3}{x-3}-x^3\,{\mathrm {e}}^{-16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x - exp(3*log(x) - 16)*(3*x^2 - 18*x + 27))/(9*x - 6*x^2 + x^3),x)

[Out]

- 3/(x - 3) - x^3*exp(-16)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 12, normalized size = 0.50 \begin {gather*} - \frac {x^{3}}{e^{16}} - \frac {3}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2+18*x-27)*exp(3*ln(x)-16)+3*x)/(x**3-6*x**2+9*x),x)

[Out]

-x**3*exp(-16) - 3/(x - 3)

________________________________________________________________________________________